Wave Optics
- Two slits in Young’s experiment have widths in the ratio 1 : 25. The ratio of intensity at the maxima and minima in the interference pattern,
Imax is : Imin
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The ratio of slits width
= 1 (given) 25 ∴ I1 = 25 I2 1 I ∝ A2 ⇒ I1 = A12 = 25 I2 A22 1 or A1 = 5 A2 1 Amax = A1 + A2 Amin A1 - A2 = 5 + 1 = 6 = 3 5 - 1 4 2 = 
3 
2 2 = 9 4 Correct Option: D
The ratio of slits width
= 1 (given) 25 ∴ I1 = 25 I2 1 I ∝ A2 ⇒ I1 = A12 = 25 I2 A22 1 or A1 = 5 A2 1 Amax = A1 + A2 Amin A1 - A2 = 5 + 1 = 6 = 3 5 - 1 4 2 = 3 2 2 = 9 4
- In the Young’s double-slit experiment, the intensity of light at a point on the screen where the path difference is λ is K, (λ being the wave length of light used). The intensity at a point where the path difference is λ /4, will be:
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For path difference λ, phase difference = 2π rad. For path difference λ/4, phase difference
= π rad. 2
As K = 4I0 so intensity at given point where path difference isλ 4 K' = 4I0 cos2 π cos π = cos45° 4 4 = 2I0 = K 2 Correct Option: C
For path difference λ, phase difference = 2π rad. For path difference λ/4, phase difference
= π rad. 2
As K = 4I0 so intensity at given point where path difference isλ 4 K' = 4I0 cos2 π cos π = cos45° 4 4 = 2I0 = K 2
- In Young’s double slit experiment, the slits are 2 mm apart and are illuminated by photons of two wavelengths λ 1 = 12000 Å and λ 2 = 10000 Å. At what minimum distance from the common central bright fringe on the screen 2 m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other ?
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∵ y = nλD d
&htere4; n1 λ1 = n2λ2
⇒ n1 × 12000 × 10–10 = n2 × 10000 × 10–10
or, n (12000 × 10–10) = (n + 1) (10000 × 10–10)
⇒ n = 5
(∵ λ1 = 12000 × 10–10m; λ2 = 10000 × 10–10 m)
Hence, ycommon = nλ1D d = 5(12000 × 10–10)× 2 2 × 10–3
(∵ d = 2 mm amd D = 2mm)
= 5 × 12 × 10–4 m
= 60 × 10–4 m
= 6 × 10–3 m = 6 mmCorrect Option: A
∵ y = nλD d
&htere4; n1 λ1 = n2λ2
⇒ n1 × 12000 × 10–10 = n2 × 10000 × 10–10
or, n (12000 × 10–10) = (n + 1) (10000 × 10–10)
⇒ n = 5
(∵ λ1 = 12000 × 10–10m; λ2 = 10000 × 10–10 m)Hence, ycommon = nλ1D d = 5(12000 × 10–10)× 2 2 × 10–3
(∵ d = 2 mm amd D = 2mm)
= 5 × 12 × 10–4 m
= 60 × 10–4 m
= 6 × 10–3 m = 6 mm
- In Young’s double slit experiment the distance between the slits and the screen is doubled. The separation between the slits is reduced to half. As a result the fringe width
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Fringe width
β = λD d
From question D′ = 2Dand d' = d 2 β' = λD1 = 4β d1 Correct Option: C
Fringe width
β = λD d
From question D′ = 2Dand d' = d 2 β' = λD1 = 4β d1
- In Young’s double slit experiment carried out with light of wavelength (λ ) = 5000 Å, the distance between the slits is 0.2 mm and the screen is at 200 cm from the slits. The central maximum is at x = 0. The third maximum (taking the central maximum as zeroth maximum) will be at x equal to
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x = (n)λ D = 3 × 5000 × 10-10 × 2 d 0.2 × 10-3
= 1.5 × 10-2 m = 1.5 cmCorrect Option: B
x = (n)λ D = 3 × 5000 × 10-10 × 2 d 0.2 × 10-3
= 1.5 × 10-2 m = 1.5 cm
